Roller coaster speed

Question

A 100kg roller coaster cart starts from rest atop a 60 meter tall hill and
rolls down.
What is the speed of the coaster at the bottom of the hill if there is no
friction?
What is the speed if there is a constant 120N retarding force?

Answer 1

Hi Meg,

 

  Use the work - kinetic energy theorem here.  Gravity does work in the system in the amount of mgh, KE changes from zero to ½mv².  equate the two and solve for v, yielding v = √(2gh) = 34.3 m/s .  

 

  The second question is more problematic.  A retarding force is generally in opposite direction to the velocity, and the amount of energy lost therefore depends on the length of the path take.  If the path were "straight down" then the opposing force would remove 120N×60m = 7200 J of energy.  The net work done by by all forces would then be W_net = mgh - 7200 J = 28,080 J  and the final speed v = 23.7 m/s.  

 

   The roller coaster path isn't likely to be straight down.  The path is then longer than 60 m, but if if it is longer than some distance d where W_net = mgh - (120N)d ≥ 0 , then d ≤ 490 m where the roller coaster comes to rest.  The final speed would then be somewhere between 0 and 23.7 m/s.